> Q. A box is lying on smooth floor pushed with a force of $10, N$ acting at an angle of $60^circ$ with horizontal. Horizontal displacement of the box is $3, m$. The work done is – LIVE ANSWER TODAY

Q. A box is lying on smooth floor pushed with a force of $10, N$ acting at an angle of $60^circ$ with horizontal. Horizontal displacement of the box is $3, m$. The work done is

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A box is lying on smooth floor pushed with a force of $10, N$ acting at an angle of $60^circ$ with horizontal. Horizontal displacement of the box is $3, m$. The work done is

Solution:

Using W = FS cos $theta$, we get W = 10 $times$ 3 $times$ cos $60^circ$ = $10 times 3 times frac{1}{2}$ = 15 J

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